proc 5071: process equipment design i - mun
TRANSCRIPT
1
PROC 5071:Process Equipment Design I
Log Mean Temperature Difference (LMTD)
Salim Ahmed
Salim Ahmed PROC 5071: Process Equipment Design I
2
1 Why log mean?
•We get a general expression for heat transfer ina heat exchanger as
Q = UA∆Tm (1)
• For ∆Tm a logarithmic mean temperature isused which is commonly known as the Log MeanTemperature Difference or LMTD.
•Question arises “Why log mean?”
2 Mathematical formulation
• Let’s look at the expression for LMTD
∆Tlm =∆T1 −∆T2
ln ∆T1∆T2
(2)
• Here, ∆T1 and ∆T2 are the temperature dif-ferences between the hot and cold fluid at thetwo ends and ∆Tlm is the LMTD.
• Figure 1 shows possible temperature profiles of
Salim Ahmed PROC 5071: Process Equipment Design I
2.1 Basic heat transfer equations 3
T
t
∆T1 ∆T2
Length of Heat Exchanger
Tem
per
ature
Figure 1: Possible temperature profiles in a parallel flow heat exchanger.
the hot and cold fluids in a heat exchanger.
• Eq. 2 shows that the ∆Tlm depends only on∆T1 and ∆T2.
• Then, does it matter how the temperaturesof the two fluids change within the heat ex-changer?
•What are the assumptions in defining the LMTD?
2.1 Basic heat transfer equations
• In a heat exchanger, the driving force is thetemperature difference between the hot and thecold fluid, ∆T = T − t
Salim Ahmed PROC 5071: Process Equipment Design I
2.1 Basic heat transfer equations 4
• Here, we denote the temperatures of the hotand cold fluid by T and t, respectively.
• Along the length of the heat exchanger T andt vary significantly and so does the ∆T
• Consequently, the heat flux also varies alongthe length.
• For a differential area dA, the rate of heat flowdq can be expressed as
T
t
∆T1 ∆T2
Length of Heat Exchanger
Tem
per
ature
dA
dT
dt
Figure 2: Hot and cold fluid temperature profiles in a heat exchanger.
dq = U(dA)∆T (3)
• Here, U is the local overall heat transfer coef-ficient.
• For the same differential area the heat transfer
Salim Ahmed PROC 5071: Process Equipment Design I
2.2 The assumptions 5
by the hot and cold fluids are given by
dqh = mhCphdT (4)
dqc = mcCpcdt (5)
• Here m and Cp are the fluid mass flow rate andspecific heat capacity, respectively.
• The subscripts h and c are used for the hot andthe cold fluid.
• dT and dt are the temperature changes in thehot and cold fluids, respectively, between theinlet and outlet of the differential area.
• To get an expression for heat transfer over theentire area using a mean value of ∆T , someassumptions need to be made.
2.2 The assumptions
Four simplifying assumptions are made
1. The specific heats of the hot and cold fluids areconstant
Salim Ahmed PROC 5071: Process Equipment Design I
2.3 Implications of the assumptions 6
2. Flow of the fluids are steady and are either par-allel or countercurrent
3. Heat loss is negligible
4. The overall heat transfer coefficient is constant
2.3 Implications of the assumptions
• Assumption (3) implies that
dqh = dqc = dq (6)
• Assumptions (1) and (2) imply that both dTdqh
and dtdqc
are constant
• Assumption (4) imply thatd(∆T )dq is constant
The above imply that T , t and ∆T are linearfunctions of q.
2.4 Expression for the LMTD
• If the total heat transfer in a heat exchanger isgiven by Q, as ∆T varies linearly with q, we
Salim Ahmed PROC 5071: Process Equipment Design I
2.4 Expression for the LMTD 7
T
t
∆T1 ∆T2
∆T
q
Tem
per
atu
re
Figure 3: Temperature versus heat flow in a parallel flow heat exchanger.
getd(∆T )
dq=
∆T2 −∆T1
Q(7)
• Using Eq. 3 we get
d(∆T )
UdA∆T=
∆T2 −∆T1
Q(8)
• A simple rearrangement gives
d(∆T )
∆T=
U(∆T2 −∆T1)
QdA (9)
• Now Q is the heat transferred over the entirearea, A. So integrating over the area between
Salim Ahmed PROC 5071: Process Equipment Design I
2.4 Expression for the LMTD 8
the two ends 1 and 2 gives∫ ∆T2
∆T1
d(∆T )
∆T=
∫ A
0
U(∆T2 −∆T1)
QdA
(10)
• Integration and use of the limits will give
ln∆T2
∆T1=
U(∆T2 −∆T1)
QA (11)
• So we get
Q = UA∆T2 −∆T1
ln ∆T2∆T1
(12)
• Comparing with Eq. 1, we get the expressionfor the mean temperature, which is known asLMTD
∆Tlm =∆T2 −∆T1
ln ∆T2∆T1
=∆T1 −∆T2
ln ∆T1∆T2
(13)
Salim Ahmed PROC 5071: Process Equipment Design I
2.5 Applicability and limitations 9
2.5 Applicability and limitations
Note that the assumptions mentioned above shouldbe satisfied for the Eq. 16 to be applicable. TheLMTD measure is not applicable when
•∆T1 and ∆T2 are equal or nearly equal. In sucha case the arithmetic mean can be used.
• U changes significantly
•∆T is not a linear function of q
2.6 Special case with variable U
•When U is not constant, if it changes linearlywith ∆T over the entire range of the heat ex-changer, a log mean of the entire term U∆Tcan be used
• In that case, one can write
Q = A(U∆T )lm (14)
Salim Ahmed PROC 5071: Process Equipment Design I
10
• with
(U∆T )lm =U2∆T1 − U1∆T2
ln U2∆T1U1∆T2
(15)
• Note that in this equation the multiplicationterm contains U of one end with ∆T of theother end
• The derivation is shown in the Appendix sec-tion.
3 Workbook: LMTD calculation for par-
allel and countercurrent flow
3.1 The problem
Methanol condensate is to be subcooled from 95oCto 50oC. Water will be used as the coolant,with a temperature rise from 25oC to 40oC. If adouble pipe heat exchanger is used, calculate theLMTD for both parallel flow and countercurrentflow arrangement.
Salim Ahmed PROC 5071: Process Equipment Design I
3.2 Notes and analysis 11
3.2 Notes and analysis
• The expression for LMTD is given by
∆Tlm =∆T1 −∆T2
ln ∆T1∆T2
(16)
• Note that in this expression 1 and 2 denote thetwo ends of the heat exchanger.
• So for parallel and countercurrent flow the def-initions of ∆T1 and ∆T2 will be different.
•We will denote the inlet end of the hot streamas end 1. Also we will use T for the hot streamand t for the cold stream.
3.3 Parallel flow
Step 1 : Identify the temperatures.
Salim Ahmed PROC 5071: Process Equipment Design I
3.3 Parallel flow 12
Using the above notations, for the parallel flow
T1 = 95oC
T2 = 50oC
t1 = 25oC
t2 = 40oC
Step 2 : Calculate ∆T1 and ∆T2.
∆T1 = T1 − t1= 95oC − 25oC
= 70oC
∆T2 = T2 − t2= 50oC − 40oC
= 10oC
Step 3 : Calculate ∆Tlm.
∆Tlm =(70− 10)oC
ln 70oC10oC
= 30.8oC
Salim Ahmed PROC 5071: Process Equipment Design I
3.4 Countercurrent flow 13
3.4 Countercurrent flow
• Note that only Step 1 in this approach is dif-ferent for parallel and countercurrent flow.
•With end 1 being the inlet for the hot stream,it’s the outlet for the cold stream in the coun-tercurrent setting.
•Once T1, T2 and t1, t2 are defined, Step 2 and3 remain the same.
Step 1 : Identify the temperatures.
Using the notations described earlier, for the coun-tercurrent flow
T1 = 95oC
T2 = 50oC
t1 = 40oC
t2 = 25oC
Step 2 : Calculate ∆T1 and ∆T2.
Salim Ahmed PROC 5071: Process Equipment Design I
3.5 Comment 14
∆T1 = T1 − t1= 95oC − 40oC
= 55oC
∆T2 = T2 − t2= 50oC − 25oC
= 25oC
Step 3 : Calculate ∆Tlm.
∆Tlm =(55− 25)oC
ln 55oC25oC
= 38oC
3.5 Comment
• Note that the LMTD is significantly higher forthe countercurrent flow than that for the par-allel flow.
• For a given set of temperatures, that is alwaysthe case.
Salim Ahmed PROC 5071: Process Equipment Design I
15
4 LMTD calculation for multi-pass shell
and tube heat exchangers
T
t
Length of Heat Exchanger
Tem
perature
Figure 4: Temperature profiles in a 1-2 shell and tube heat exchanger.
• For multipass heat exchangers, the temperatureprofiles become complex.
• The concept of inlet end and outlet end becomeinvalid.
• For multipass heat exchangers, an approximatemethod is used to calculate the mean temper-ature difference.
Salim Ahmed PROC 5071: Process Equipment Design I
4.1 Calculation steps 16
4.1 Calculation steps
Step 1 : LMTD (∆lm) is calculated assuming asingle pass countercurrent flow.
Step 2 : A correction factor, FT , is calculatedbase on the temperatures of the hot and coldstream and the type of heat exchanger.
Step 3 : The corrected mean temperatures iscalculated as.
∆Tm = FT ×∆lm (17)
5 Workbook: LMTD calculation for a 1-
2 shell and tube heat exchanger
5.1 The problem
Methanol condensate is to be subcooled from 95oCto 50oC. Water will be used as the coolant, witha temperature rise from 25oC to 40oC. If a 1-2shell and tube heat exchanger is used, calculate
Salim Ahmed PROC 5071: Process Equipment Design I
5.2 Notes and analysis 17
the corrected mean temperature.
5.2 Notes and analysis
•We will use the notations T and t for the hotand cold fluid temperature, respectively.
• For the ends, 1 is used for the inlet of the hotstream.
• For countercurrent, end 2 is the inlet for thecold fluid.
5.3 Calculation steps
Step 1 : Calculate LMTD (∆lm) assuming a sin-gle pass countercurrent flow.
1.1 : Identify the temperatures.
Using the above mentioned notations, for the coun-
Salim Ahmed PROC 5071: Process Equipment Design I
5.3 Calculation steps 18
tercurrent flow
T1 = 95oC
T2 = 50oC
t1 = 40oC
t2 = 25oC
1.2 : Calculate ∆T1 and ∆T2.
∆T1 = T1 − t1= 95oC − 40oC
= 55oC
∆T2 = T2 − t2= 50oC − 25oC
= 25oC
1.3 : Calculate ∆Tlm.
∆Tlm =(55− 25)oC
ln 55oC25oC
= 38oC
Salim Ahmed PROC 5071: Process Equipment Design I
5.3 Calculation steps 19
Step 2 : Calculate the correction factor, FT .
2.1 : Calculate the constants R and S
R is defined as the ratio of temperature decreaseof the hot fluid to the temperature increase of thecold fluid
R =T1 − T2
t2 − t1
=(95− 50)oC
(40− 25)oC= 3.0
S is defined as the ratio of the temperature in-crease of the cold fluid to the difference in theinlet temperature of the two fluids
S =t2 − t1T1 − t2
=(40− 25)oC
(95− 25)oC= 0.21
2.2 : From the graph find the value of the cor-
Salim Ahmed PROC 5071: Process Equipment Design I
5.3 Calculation steps 20
rection factor.
Figure 5: Temperature correction factor for 1-2 shell and tube heat exchangers.
Alternatively, the following equation can be usedto calculate FT .
FT =
√R2 + 1 ln
[1−S
1−RS
](R− 1) ln
[2−S
(R+1−
√R2+1
)2−S
(R+1+
√R2+1
)]Step 3 : Calculate the corrected mean tempera-
ture.
Salim Ahmed PROC 5071: Process Equipment Design I
21
∆Tm = FT ×∆Tlm= 0.94× 38oC
= 35.7oC
6 Use of FT for selecting heat exchanger
configuration
• The configuration of the heat exchanger (num-ber of shell and tube passes) is selected to geta desired FT .
• It is desirable to have FT > 0.85.
• FT < 0.75 is generally unacceptable.
• For a given number of shell pass, the value ofFT is not affected significantly on the numberof tube passes.
• The more shell passes, the higher is the valueof FT .
Salim Ahmed PROC 5071: Process Equipment Design I
22
Appendix
• This section provides the derivation of the heattransfer equation when U is not constant andchanges linearly with ∆T
•WE start with Eq. 8
d(∆T )
UdA∆T=
∆T2 −∆T1
Q(18)
•We rearrange the equation considering that Uis a function of ∆T
d(∆T )
U∆T=
∆T2 −∆T1
QdA (19)
• As U changes linearly with ∆T , we can expressU as
U = a + b∆T (20)
• As U1 = a + b∆T1 and U2 = a + b∆T2
a =U2∆T1 − U1∆T2
∆T2 −∆T1
b =U2 − U1
∆T2 −∆T1
Salim Ahmed PROC 5071: Process Equipment Design I
23
• Integrating the equation between the limits atthe two ends∫ ∆T1
∆T1
d(∆T )
(a + b∆T )∆T=
∫ A
0
U(∆T2 −∆T1)
QdA
(21)
•We will use the following formula∫dx
(a + bx)x=
1
aln
x
a + bx(22)
• Upon integration, we get left hand side (LHS)of the equation as
LHS =1
aln
∆T
a + b∆T
∣∣∣∣∆T1
∆T1
=1
a
[ln
∆T2
a + b∆T2− ln
∆T1
a + b∆T1
]=
∆T2 −∆T1
U2∆T1 − U1∆T2lnU1∆T2
U2∆T1(23)
Salim Ahmed PROC 5071: Process Equipment Design I
24
• So we have
∆T2 −∆T1
U2∆T1 − U1∆T2lnU1∆T2
U2∆T1=
∆T2 −∆T1
QA
(24)
• Upon rearrangement, one get
Q = AU2∆T1 − U1∆T2
ln U2∆T1U1∆T2
(25)
• Comparing Eq. 25 with Eq. 14, we get the ex-pression of (U∆)lm as in Eq. 16.
Salim Ahmed PROC 5071: Process Equipment Design I
25
References
1. W. L. McCabe, J. C. Smith, P. Harriott. (2005).Unit Operations of Chemical Engineering, 7thEdition, McGraw Hill, New York, USA.
2. G. Towler and R. Sinnott. Chemical Engineer-ing Design: Principles, Practice and Economicsof Plant and Process Design. Butterworth-Heinemann.2008.
Salim Ahmed PROC 5071: Process Equipment Design I